The cute little thing about a string of 9’s followed by a 6 being a number divisible by 6 inspired my Dearly Beloved, who spent some time looking for other patterns in this kind of number. I’m glad for that; this sort of pattern, while it may not be terribly important, is often fun to play with. And interesting things can be found in play.
I don’t know a good name for this kind of number, and admit it feels awkward to say just “this kind of number”. If I have to talk about them much longer some group name is probably worth devising. Unfortunately the only names which come to my mind come there through organic chemistry, where it’s reasonably common to have an arbitrarily long chain of carbon atoms terminated with some distinctly different group. For example, an alcohol is a string of carbons ending with an oxygen and hydrogen molecule. But an “alcoholic number”, while an imagination-capturing name, doesn’t quite fit. I suppose aldehydes, which end on a double-bond to an oxygen atom, preserves the metaphor, but no one knows the adjective form of aldehyde.
My Dearly Beloved’s experiments found no other numbers for which a repeated string, terminated by a 6, would produce a number divisible by 6. This overlooked the obvious case, though: a string of 6’s, followed by another 6, is itself divisible by 6. Obvious cases are like that, and many people would think of a uniform string of 6’s not part of the pattern “an arbitrary number of one digit, followed by a 6”.
But there is another case, which I don’t doubt my Dearly Beloved would have noticed if not distracted into thinking of whether … well, let’s give in to calling them alcoholic numbers, for a few hundred words at least … might be divisible by other digits. That case is the string of 3’s, terminated by 6. A proof of this is just like one for the case of 9’s followed by a 6, and we can copy last week’s with almost no modification. 36 minus 6 is 30, which is divisible by 6. 336 minus 36 is 300, which is divisible by 6. 3,336 minus 336 is 3,000, and so on.
The point of the discovery holds, though: arbitrary strings of 1’s, 2’s, 4’s, 5’s, 7’s, or 8’s ending in a 6 don’t produce a number divisible by 6. What is there different about those digits?
We’ll come to that, but I want first to look at the other thing to explore. Can we produce such alcoholic numbers which are divisible by other numbers? Divisibility by 5 is easy: let anything be the repeating digit, and as long as the terminating digit is 5 or 0, the result is divisible by 5. There’s a similar universality to getting numbers divisible by 2: anything with an even terminating digit is divisible by 2. And everything is divisible by 1, so there isn’t any point looking for patterns. Any repeating and any terminating digit will do.
What about 3? A repeating digit of 1 and terminating digit of 3 doesn’t work, as 13 shows. Repeating 2 doesn’t work, but repeating 3 does. 4 and 5 fail, but 6 works, and now we start looking suspiciously at these alcoholic numbers, for giving divisibility by 3 and divisibility by 6 for the same repeating digits. Of course, a repeating digit of 3, 6, or 9 terminated with a 6 is also divisible by 3; similarly, if the terminating digit is 9 or 0 we have something divisible by 3.
But if we want a number terminating in 2 divisible by 3 … a repeating 1 gives us hope in the first case: 12 certainly works. But 112 does not; and 1,112 does not; while 11,112 does again. 111,112 is not divisible by 3; 1,111,112 is not; 11,111,112 is. We can build similar patterns of sometimes-divisible, sometimes-not numbers terminating in 1 or any of the other digits. And looking back to the divisibility by 6 … 86 and 886 gave no reason to expect there was a pattern there, but 8,886 divided by 6 is the neat little 1481. There’s something further to explore there.
For divisibility by 4 we get another an interesting split: if the repeating digit is even and the terminating digit 0, 4, or 8, the number formed is divisible by 4. If the repeating digit is odd, the terminating digit must be 2 or 6 for the result to be divisible by 4. There is none of the some-work, some-fail case that we see for divisibility by 3. Why should that happen?
Are there any alcoholic numbers divisible by 7 and 8, other than the strings of nothing but 7, and 4’s or 8’s before the 8?
We’re run into something interesting here.
nebusresearch 
It seems to me that you have missed some of the main divisibility tests, particularly the one which says any number whose digits sum to a multiple of 3 is divisible by 3. The same for 9. And any number divisible by 9 is divisible by 3. So any such number which is also even is divisible by 3 and 2, hence divisible by 6. Which is why 9…96 is always divisible by 3 and 6.
[to take an example of a 4 digit number wxyz = 1000w + 100x + 10y + z = 9(111w + 11x + y) + (w + x + y + z), so if w+x+y+z is a multiple of 9, the whole expression is]
Any number with the last two digits divisible by 4 is divisible by 4 (because 100 is divisible by 4). It’s the last three digits for divisibility by 8, last 4 for 16, and so on.
And any number where the two sets of alternate digits sum to the same value, or to values with a difference of 11 or some multiple thereof, are divisible by 11.
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Oh, no, none of that missed at all. I’m just building to them in a slightly roundabout way.
(Sorry to be so long in following up. I was visiting my Dearly Beloved over the weekend and that took priority over such things as this project.)
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Ah, to my mind you’re going round the world to the West in order to travel to you neighbour to the East, more than slightly roundabout. I am rather bemused. Never mind.
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I am indeed taking about the longest way around I can; I’d figured it would let me avoid being stuck for topics. This is going to seem really hilarious to me come Monday.
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Your divisibility by 3 could be cleaner. By the rules of modular arithmetic, if a1 ≡ b1 (mod n) and a2 ≡ b2 (mod n) then a1 + a 2 ≡ b1 + b2 (mod n). Additionally if a and b ∈ Z then a1a2 ≡ a1b2. This makes all positive integer powers of 10 ≡ 1 (mod 3). By the multiplication rule, this makes all the powers of 10 drop out, leaving us with just, in your example w + x + y + z. The rest is just the application of the rule for addition.
At leat I think it is. I could be wrong.
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In fact I was wrong. Also WordPress doesn’t seem to allow %lt;sub> but does do entities. Who knew. To correct then:
if a_1 ≡ b_1 (mod n) and a_2 ≡ b_2 (mod n) then a_1 + a_2 ≡ b_1 + b_2 (mod n) :: Addition rule
if a and b are both integers, then a_1 a_2 ≡ b_1 b_2 (mod n) :: Muliplication rule.
10^0 = 1 ≡ 1 (mod 3); 10^1 = 10 &equiv 1 (mod 3). The rest of those follow by the multiplication rule.
I think the rest is fairly solid. But again, reserve the right to be wrong.
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A string of threes followed by a six is always divisible by six.
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It is indeed. Works out nicely that way since 30 is also divisible by six. (At least that’s one way to go proving it.)
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